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Tambaya 1 Rahoto
Which of the following will give a precipitate with an aqueous solution of copper (I) chloride?
Bayanin Amsa
Tambaya 2 Rahoto
Hydrogen diffused through a porous plug
Bayanin Amsa
Hydrogen gas (H2) diffuses faster than oxygen gas (O2) through a porous plug. This is because the rate of diffusion of a gas through a porous plug is inversely proportional to the square root of its molar mass. Since the molar mass of hydrogen (2 g/mol) is much smaller than that of oxygen (32 g/mol), the rate of diffusion of hydrogen through a porous plug is much faster than that of oxygen. To be more specific, the ratio of the diffusion rates of two gases through a porous plug is given by the equation: Rate of diffusion of gas A / Rate of diffusion of gas B = √(Molar mass of gas B / Molar mass of gas A) Using the molar masses of hydrogen and oxygen, we get: Rate of diffusion of hydrogen / Rate of diffusion of oxygen = √(32 g/mol / 2 g/mol) = √16 = 4 Therefore, hydrogen diffuses through a porous plug four times as fast as oxygen. Thus, the correct answer is: four times as fast as oxygen.
Tambaya 4 Rahoto
2-methylprop-1-ene is an isomer of
Bayanin Amsa
2-methylprop-1-ene is an isomer of 3-methyl but-1-ene and 2-methyl but-1-ene. An isomer is a molecule that has the same molecular formula as another molecule, but a different arrangement of atoms. In this case, 2-methylprop-1-ene has the molecular formula C4H8, and so do 3-methyl but-1-ene and 2-methyl but-1-ene. The difference between these three molecules is in the arrangement of the carbon and hydrogen atoms. 2-methylprop-1-ene has a branched structure with a double bond between the first and second carbon atoms. 3-methyl but-1-ene is also a branched molecule, but the double bond is between the second and third carbon atoms. Similarly, 2-methyl but-1-ene has a double bond between the first and second carbon atoms, but it has a different branching pattern. On the other hand, pent-2-ene has five carbon atoms, so it has a different molecular formula than 2-methylprop-1-ene. Therefore, 2-methylprop-1-ene is an isomer of 3-methyl but-1-ene and 2-methyl but-1-ene, but not of pent-2-ene, because it has the same molecular formula and a different arrangement of atoms compared to the other two isomers.
Tambaya 5 Rahoto
X is a substance which liberates CO2 on treatment with concentrated H2 SO4 . A warm solution of X can decolorize acidified KMnO4 . X is
Bayanin Amsa
It should be noted that for X to liberate CO2
, X must be a carbonate or an oxalate. Since X decolorizes KMnO4
, X must be an oxalate.
Therefore, X is H2
C2
O4
.
Tambaya 6 Rahoto
Na2 CO3 + 2HCl → 2NaCl + H2 O + CO2
The indicator most suitable for this reaction should have a pH equal to
Bayanin Amsa
Methyl orange is the best indicator for the reaction with range 3.1 - 4.4.
Tambaya 7 Rahoto
What mass of magnesium would be obtained by passing a current of 2 amperes for 2 hours, through molten magnesium chloride?
[1 faraday = 96500C, Mg = 24]
Bayanin Amsa
Current (I) = 2A; Time (t) = 2 hours = 7200 secs
Q = It
= 2 x 7200 = 14400C
1 F = 96500C
x = 14400C
x = 1440096500
= 0.15F
Mg2+
+ 2e−
→
Mg
2F →
24g
0.15F →
x
2x = 24 x 0.15
x = 24×0.152
= 1.8g
Tambaya 8 Rahoto
Hydrocarbons which will react with Tollen's reagent conform to the general formula
Tambaya 9 Rahoto
The heat of formation of ethene, C2 H4 is 50 kJmol−1 , and that of ethane, C2 H6 is -82kJmol−1 . Calculate the heat evolved in the process:
C2 H4 + H2 → C2 H6
Bayanin Amsa
The heat evolved in a chemical reaction can be calculated by subtracting the heat of formation of the reactants from the heat of formation of the products. In this case, the reactants are ethene (C2H4) and hydrogen (H2), and the product is ethane (C2H6). The heat of formation of ethene is 50 kJ/mol and that of hydrogen is 0 kJ/mol (because hydrogen is a reference element). The heat of formation of ethane is -82 kJ/mol. So, the heat evolved in the reaction is given by: Heat evolved = (Heat of formation of products) - (Heat of formation of reactants) = (-82 kJ/mol) - (50 kJ/mol + 0 kJ/mol) = -82 kJ/mol - 50 kJ/mol = -132 kJ/mol. Therefore, the heat evolved in the process is -132 kJ.
Tambaya 10 Rahoto
When chlorine water is exposed to bright sunlight, the following products are formed
Tambaya 11 Rahoto
Which of the following conditions will most enhance the spontaneity of a reaction?
Bayanin Amsa
The condition that will most enhance the spontaneity of a reaction is when ΔH is negative (i.e., the reaction releases heat) and ΔS is positive (i.e., the reaction increases the disorder or randomness of the system). This is because a negative ΔH indicates that the reaction releases energy, which is favorable for a spontaneous reaction, while a positive ΔS indicates that the system becomes more disordered, which is also favorable for spontaneous reactions. Among the given options, the first condition of a negative and greater ΔH than ΔS is the best option for enhancing the spontaneity of a reaction. The other options have either a positive ΔH or a zero ΔS, which is not favorable for spontaneous reactions.
Tambaya 12 Rahoto
When ammonia and hydrogen ion bond together to form ammonium ion, the bond formed is called
Bayanin Amsa
When ammonia and hydrogen ion go into bonding, they form ammonium ion by combining with a dative/coordinate covalent bond.
Tambaya 13 Rahoto
Which of the following could not be alkane?
Bayanin Amsa
An alkane is a type of hydrocarbon with only single bonds between the carbon atoms. It follows the general formula CnH2n+2, where "n" is the number of carbon atoms in the molecule. To determine whether a molecule is an alkane or not, we can calculate its molecular formula and check if it fits the general formula of alkane. Out of the given options, the third one (C7H14) cannot be an alkane. To see why, let's use the general formula of alkane, which is CnH2n+2. For C7H14 to be an alkane, it should have 2n+2 = 2(7) + 2 = 16 hydrogen atoms. However, C7H14 has only 14 hydrogen atoms, which means it does not follow the general formula of alkane. Therefore, C7H14 cannot be an alkane. The other options are as follows: - C4H10: This is butane, which is an alkane with four carbon atoms. - C5H12: This is pentane, which is an alkane with five carbon atoms. - C8H18: This is octane, which is an alkane with eight carbon atoms. In summary, the molecule C7H14 cannot be an alkane because it does not follow the general formula of alkane, while the other options are all examples of alkanes.
Tambaya 14 Rahoto
The cost of discharging 6.0g of a divalent metal, X from its salt is ₦12.00. What is the cost of discharging 9.0g of a trivalent metal, Y from its salt under the same condition?
[X = 63, Y = 27, 1F = 96,500C]
Bayanin Amsa
For X: X2+
+ 2e−
→
X
2F = 63g
xF = 6g
x = 6×263=421F
421
F = N12.00
1F = 12421
= N63.00
1F is equivalent to N63.00.
For Y: Y3+
+ 3e−
→
Y
3F = 27g
xF = 9g
x = 3×927
= 1F
1F = N63.00
Tambaya 16 Rahoto
By what amount must the temperature of 200cm3 of Nitrogen at 27°C be increased to double the pressure if the final volume is 150cm3 (Assume ideality)
Bayanin Amsa
Using the ideal gas law and equation:
P1V1T1=P2V2T2
P1×200cm3300K=2P×150cm3T2
Cross multiply:
T2=300×150×2P200×P
=450K
or 177∘C
Don't forget to convert to ∘C
Tambaya 18 Rahoto
In the reaction:
M + N → P
ΔH = +Q kJWhich of the following would increase the concentration of the product?
Bayanin Amsa
Increasing the temperature would increase the concentration of the product, P. The reaction rate, or the speed at which the reaction occurs, is influenced by temperature. An increase in temperature raises the kinetic energy of the reacting molecules, making it easier for them to collide and react. This leads to a higher rate of reaction and a higher concentration of the product, P. Adding a suitable catalyst can also increase the reaction rate, but it does not directly affect the concentration of the product. Increasing the concentration of P does not affect the reaction itself, but is a result of the reaction having taken place. Decreasing the temperature would slow down the reaction rate and reduce the concentration of the product.
Tambaya 19 Rahoto
Which process(es) is/are involved in the turning of starch iodide paper blue-black by chlorine gas?
Bayanin Amsa
The process involved in the turning of starch iodide paper blue-black by chlorine gas is option number 3: chlorine oxidizes the iodide ion to produce iodine which attacks the starch to give the blue-black color. When chlorine gas comes in contact with iodide ions on the starch iodide paper, it oxidizes the iodide ion to form iodine. The iodine that is produced in this reaction is then able to react with the starch present on the paper to form a blue-black complex. This blue-black complex is formed due to the arrangement of the starch molecules and the iodine atoms in a way that causes them to absorb light at a specific wavelength, giving the blue-black color. Therefore, the blue-black color that is observed on the starch iodide paper is due to the reaction between iodine and starch, which is made possible by the oxidation of iodide ions by chlorine gas.
Tambaya 20 Rahoto
Consider the reaction: A + 2B(g)⇌ 2C + D(g) (Δ H = +ve)
What will be the effect of decrease in temperature on the reaction?
Bayanin Amsa
The effect of a decrease in temperature on the reaction will be that the rate of the backward reaction will increase. In a chemical reaction, the rate of the forward and backward reactions are determined by the activation energy required for each step and the temperature of the system. When the temperature is decreased, the rate of the reaction decreases, and the rate of the backward reaction increases. This shift in the rate of the backward reaction means that there will be a shift in the position of the equilibrium of the reaction. As the rate of the backward reaction increases, the concentration of the reactants will increase and the concentration of the products will decrease, leading to a decrease in the overall yield of the products. In this reaction, as ΔH (the change in enthalpy) is positive, which means that the reaction is endothermic. Endothermic reactions absorb heat from the surroundings to proceed, so a decrease in temperature will lead to a decrease in the rate of the forward reaction and an increase in the rate of the backward reaction. This shift in the rate of the backward reaction will shift the position of the equilibrium of the reaction to the left, leading to an increase in the concentration of the reactants and a decrease in the concentration of the products.
Tambaya 21 Rahoto
The IUPAC nomenclature of the compound
H3 C - CH(CH3 ) - CH(CH3 ) - CH2 - CH3
Tambaya 22 Rahoto
A certain hydrocarbon on complete combustion at s.t.p produced 89.6dm3 of CO2 and 54g of water. The hydrocarbon should be
Bayanin Amsa
In the question above an Hydrocarbon combust to give CO2 and H20
Let Hydrocarbon be
CxHy + x+Y/4O2= xCO2 + Y/2H2O
Mass of C0=44g and H2O=18g
at STP vol= 22.4
Therefore, 1mole of CO2 contains 44g and 22.4dm³ at STP
1mole = 22.4dm³
xmole = 89.6dm³
Cross multiplying x=89.6/22.4 =4mole of CO2 produce
1mole of H2O = 18g
Xmole = 56g
Cross multiplying
X = 56/18 = 3mole of H20
Then....
CxHy + X + y/4O2 = 4CO2+ 3H2O
Balancing
C4H6 + 6O2 = 4CO2 + 3H2O
Tambaya 23 Rahoto
Consider the equation below:
Cr2 O2−7 + 6Fe2+ + 14H+ → 2Cr3+ + 6Fe3+ + 7H2 O.
The oxidation number of chromium changes from
Bayanin Amsa
Cr2
O2−7
+ 6Fe2+
+ 14H+
→
2Cr3+
+ 6Fe3+
+ 7H2
O
The oxidation of Cr in Cr2
O2−7
:
Let the oxidation of Cr = x;
2x + (-2 x 7) = -2 ⟹
2x - 14 = -2
2x = 12 ; x = +6
Hence, the change in oxidation of Cr = +6 to +3
Tambaya 24 Rahoto
The velocity, V of a gas is related to its mass, M by (k = proportionality constant)
Bayanin Amsa
Recall:
V = √3RTM
∴V∝1√M
V=k√M
V = kM12
Tambaya 25 Rahoto
At 27°C, 58.5g of sodium chloride is present in 250cm3 of a solution. The solubility of sodium chloride at this temperature is?
(molar mass of sodium chloride = 111.0gmol−1 )
Bayanin Amsa
Given the Mass of the salt = 58.5g
Volume = 250 cm3
= 0.25 dm3
Mass concentration = MassVolume
= 58.50.25
= 234 gdm−3
Solubility (in moldm−3
= 234111
= 2.11 moldm−3
≊
2.0 moldm−3
Tambaya 26 Rahoto
A solution X, on mixing with AgNO3 solution gives a white precipitate soluble in aqueous NH3 , a solution Y, when also added to X, also gives a white precipitate which is soluble when heated solutions X and Y respectively contain
Tambaya 27 Rahoto
Which quantum divides shells into orbitals?
Bayanin Amsa
The quantum that divides shells into orbitals is the "Azimuthal" quantum number, also known as the "angular momentum" quantum number. The azimuthal quantum number determines the shape of an electron's orbital, which is a region in space where there is a high probability of finding an electron. It describes the angular momentum of an electron in an atom and the number of subshells within a given shell. Each subshell is associated with a specific shape, and can hold a certain number of electrons. The azimuthal quantum number is represented by the letter "l" and can have integer values ranging from 0 to (n-1), where "n" is the principal quantum number. Each value of "l" corresponds to a different subshell shape: - l = 0 corresponds to an "s" subshell, which is spherical in shape. - l = 1 corresponds to a "p" subshell, which has a dumbbell shape with two lobes. - l = 2 corresponds to a "d" subshell, which has a more complex shape with four lobes and a doughnut-like ring. - l = 3 corresponds to an "f" subshell, which has an even more complex shape with eight lobes. The number of orbitals within a subshell is equal to 2l+1. For example, a "p" subshell (l = 1) has three orbitals (2l+1 = 3), which are labeled as "px", "py", and "pz". In summary, the azimuthal quantum number determines the shape of the electron's orbital and the number of subshells within a given shell, and it is represented by the letter "l".
Tambaya 28 Rahoto
Which of the following properties increases from left to right along the period but decreases down the group in the Periodic Table?
I. Atomic Number ii. Ionization energy iii. Metallic character iv. Electron affinity
Bayanin Amsa
Ionization energy and electron affinity increase across a period, and decrease down a group.
Tambaya 29 Rahoto
SO3 is not directly dissolved in water in the industrial preparation of H2 SO4 by the contact process because
Bayanin Amsa
Tambaya 30 Rahoto
How many electrons will be found in the nucleus of an atom with mass number 23 and 17 neutrons?
Bayanin Amsa
Electrons are not found in the nucleus of an atom. The nucleus of an atom only contains protons and neutrons, while electrons are located outside the nucleus in the electron cloud. The mass number of an atom is equal to the sum of the number of protons and the number of neutrons in the nucleus. Therefore, if an atom has a mass number of 23 and 17 neutrons, then the number of protons in the nucleus can be calculated as: Protons = Mass number - Neutrons Protons = 23 - 17 Protons = 6 This means that the nucleus of the atom contains 6 protons. The number of electrons in a neutral atom is equal to the number of protons, so the atom also contains 6 electrons in the electron cloud surrounding the nucleus. In summary, the answer is that there are 6 protons and 6 electrons in the atom.
Tambaya 31 Rahoto
The IUPAC name of the compound CF3 CHBrCl is
Tambaya 32 Rahoto
A radioactive nucleus has a half-life of 20 years, starting with 100,000 particles, how many particles will be left exactly at the end of 40 years
Bayanin Amsa
The half-life of a radioactive nucleus is the time it takes for half of its particles to decay. This means that after 20 years, 100,000 particles will become 50,000 particles. After 40 years, we can find the number of particles remaining by counting the number of half-lives that have passed. Since 40 years is double the half-life of 20 years, this means that two half-lives have passed, so the number of particles will be halved twice. Starting with 100,000 particles: - After 1 half-life (20 years), there will be 50,000 particles remaining. - After 2 half-lives (40 years), there will be 25,000 particles remaining. So, exactly at the end of 40 years, there will be 25,000 particles remaining.
Tambaya 33 Rahoto
What volume of 0.100M sodium trioxonitrate (V) solution contains 5g of solute.
[Na = 23, N = 14, O = 16]
Bayanin Amsa
To calculate the volume of a solution, we need to use the formula: moles of solute = concentration x volume First, let's find the number of moles of sodium trioxonitrate (V) in 5g of the solute. The molar mass of NaNO3 is: Na = 23 N = 14 3 x O = 3 x 16 = 48 Molar mass = 23 + 14 + 48 = 85 g/mol The number of moles of NaNO3 in 5g is: moles = mass / molar mass = 5 / 85 = 0.0588 moles Now, we can use the formula above to find the volume of the solution: moles of solute = concentration x volume volume = moles of solute / concentration volume = 0.0588 moles / 0.100 M volume = 0.588 litres Therefore, the correct answer is 0.588 litres of 0.100M sodium trioxonitrate (V) solution contains 5g of solute.
Tambaya 34 Rahoto
The combustion of carbon(ii)oxide in oxygen can be represented by equation.
2CO + O2 ? 2CO2
Calculate the volume of the resulting mixture at the end of the reaction if 50cm3 of carbon(ii)oxide was exploded in 100cm3 of oxygen
Bayanin Amsa
Tambaya 35 Rahoto
Burning magnesium ribbon in air removes which of the following
(i) oxygen (ii) nitrogen (iii) argon and (iv) carbon(iv)oxide?
Bayanin Amsa
Burning magnesium ribbon in air will remove oxygen (option i) from the air, but not nitrogen (option ii), argon (option iii), or carbon dioxide (option iv). When magnesium burns, it reacts with oxygen in the air to form magnesium oxide. The reaction can be represented by the following equation: 2Mg(s) + O2(g) → 2MgO(s) The magnesium in the ribbon combines with oxygen in the air to form solid magnesium oxide. This reaction is exothermic, which means that it releases heat and light energy. So, when magnesium ribbon is burned in air, it consumes the oxygen in the air to form magnesium oxide. However, nitrogen, argon, and carbon dioxide are not chemically reactive with magnesium, and therefore are not removed from the air by the burning of magnesium ribbon. In summary, the correct option is (i) only - burning magnesium ribbon in air removes oxygen only.
Tambaya 36 Rahoto
The IUPAC name for CH3 CH2 COOCH2 CH3 is
Bayanin Amsa
The IUPAC name for the given molecule is ethyl propanoate. To arrive at the IUPAC name, we first identify the longest continuous chain of carbon atoms, which in this case is a 4-carbon chain (propane). We then identify and name the substituent groups attached to this chain, which are a methyl group (CH3) attached to the second carbon atom and an ethoxy group (OC2H5) attached to the third carbon atom. The ethoxy group is named as an ethyl group, and the entire molecule is named as ethyl propanoate, following the standard IUPAC naming conventions for esters.
Tambaya 37 Rahoto
The emission of two successive beta particles from the nucleus 3215P will produce
Bayanin Amsa
Tambaya 38 Rahoto
The two ions responsible for hardness in water are
Bayanin Amsa
The ions responsible for hardness in water are Ca2+ and/or Mg2+. Hardness in water refers to the presence of calcium and magnesium ions, which are commonly found in natural water sources such as rivers, lakes, and groundwater. These ions can react with soap to form insoluble compounds, reducing the effectiveness of soap and causing scaling in pipes and appliances. The hardness of water is often measured in terms of the concentration of calcium and magnesium ions, expressed as calcium carbonate equivalents (CaCO3).
Tambaya 39 Rahoto
Consider the reaction
A(s) + 2B(g) → 2C(aq) + D(g)
What will be the effect of a decrease in pressure on the reaction?
Bayanin Amsa
Given: The equation below
A(s) + 2B(g) → 2C(aq) + D(g)
Since we have a higher number of moles of gaseous species on the LHS, i.e 2, a decrease in pressure will favor the forward reaction.
Tambaya 40 Rahoto
The electronic configuration of element Z is 1s2 2s2 2p6 3s2 3p1 . What is the formula of the compound formed between Z and tetraoxosulphate (VI) ion.
Bayanin Amsa
Z = 1s2
2s2
2p6
3s2
3p1
?
We have Z3+
and SO2?4
The reaction : Z3+
+ SO2?4
?
Z2
(SO4
)3
.
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